Given a quantified statement ∀n, n>0 (∃x, x>2k | x=2k+n) ( a subset of the natural numbers)
This can logically this can be deduced as valid; however, I wish to use induction.
Specifically I would like to use Mathematical Induction to show that, because the set of natural numbers are ordered and there exists no maximal integer in the set, that for any x you give me that is greater than 2k I can represent in the form 2k+n.
The statement has x, and n as discrete parameters for in the induction but instead of the typical P(n) property P to prove, it seems this statement is something like P(x,n).
I could restructure the argument so that: ∀n, n>2 n=2k+1 for some k OR ∀n,k n>0 2k+n = any natural number greater than 2 but I still have two variables.
Two things I'm getting a mental block on is, I'm trying to prove that a member of the natural numbers can be shown for incremental values of k,n but I'm not sure how to state each iteration after the base case for n+1, (I guess we can keep k static at 1?), some number will be a member for the subset of natural numbers greater than 2.
and what would my inductive form look like if I didn't want to change my original argument from something like 2k+n = x ⊂ N to something like ∀n, 2+n ⊂ N.
I wanted to keep the form 2k+n.
(What I'm working on is a problem where all numbers except 2k itself can be described using 2k±n for n>0. So for X@Y is a string of L and X,Y are of the exact same alphabet, X@Y is a member of L iff there is no permutations of X in Y, and non of Y in X. e.g. aba@aaba is valid but aab@baa or aab@aba are not valid. So there are two definite properties of L: |X|≠|Y| and either X is of the for 2k, and Y is of the form 2k±n or X is of the form 2k±n, and Y is of the form 2k.)